Question

The half-life of K-40 is $1.30\times {10}^9$year. A sample of 1.0 g of pure KCl gives activity of 160 dps. The percentage fraction of K-40 present in the sample is:

If $x$ is the answer then write your answer in $100\times x$.

Answer 1

The expression for the rate of radioactive decay is $r=\lambda \times N$

The decay constant is $\lambda =\frac{0.693}{t_{1/2}}$

$\lambda =\frac{0.693}{1.30\times {10}^9\times 365\times 24\times 3600}$

Hence, $160dps=\frac{0.693}{1.30\times {10}^9\times 365\times 24\times 3600}\times N$

$N=9.465\times {10}^{18}$

The number of moles present at time t is $\frac{9.465\times {10}^{18}}{6.023\times {10}^{23}}=1.572\times {10}^{-5}$

The number of moles present initially is $\frac{1}{74.5}=0.013$

The percentage fraction is $\frac{1.572\times {10}^{-5}}{0.013}\times 100=0.12$%